我如何过滤List<Map<String，dynamic>>，以得到一个颤振值？

Question

我有一个json数据，我正在使用List<Map<String, dynamic>> jsonDecode.加载数据。

例如，我有一个；

1. 数额= 12500
2. 口径= 24

我需要过滤子来获得成熟度值：689.19

另外，如果我有一个数值= 12500和值= 689.19，我需要得到口径，也就是24。

如何过滤List> json数据以获得一个颤振值？

[
   {"amount": "5000", "caliber": "12", "maturity":  "484.25"},
   {"amount": "5000", "caliber": "24", "maturity":  "275.67"},
   {"amount": "7500", "caliber": "12", "maturity":  "726.38"},
   {"amount": "7500", "caliber": "24", "maturity":  "413.51"},
   {"amount": "10000", "caliber": "12", "maturity":  "968.50"},
   {"amount": "10000", "caliber": "24", "maturity":  "551.35"},
   {"amount": "12500", "caliber": "12", "maturity":  "1210.63"},
   {"amount": "12500", "caliber": "24", "maturity":  "689.19"},
   {"amount": "15000", "caliber": "12", "maturity":  "1452.76"},
   {"amount": "15000", "caliber": "24", "maturity":  "827.03"},
   {"amount": "15000", "caliber": "12", "maturity":  "1694.89"},
   {"amount": "17500", "caliber": "24", "maturity":  "964.87"},
   {"amount": "17500", "caliber": "36", "maturity":  "727.66"},
   {"amount": "17500", "caliber": "48", "maturity":  "613.53"},
   {"amount": "17500", "caliber": "60", "maturity":  "548.44"},
   {"amount": "20000", "caliber": "12", "maturity":  "1937.01"},
   {"amount": "20000", "caliber": "24", "maturity":  "1102.71"}
]

更新:我终于得到了@Muldec帮助下的。

首先，我加载上面的儿子列表如下所示。然后应用@Muldec的答案，它就成功了。

List<Map> myList = List<Map>.from(jsonDecode(jsonFastCalculation) as List);
final item = myList.firstWhere((e) => e['amount'] == '12500' && e['caliber'] == '24');
print(item['maturity']);

Answer 1

根据搜索条件，可以使用列表中的firstWhere或where方法获取所需的Map元素或Map列表。

假设您绝对确信搜索条件只给您一个项(或者您只关心满足条件的第一个项)，下面是一个firstWhere代码示例

List<Map<String, dynamic>> myList = ....;

final item = myList.firstWhere((e) => e['amount'] == '12500' && e['caliber'] == '24');
print(item['maturity']);

在firstWhere方法中，根据数量和成熟度找到口径只是改变测试的一件事。